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Solutions Manual for Aircraft Performance: An Engineering Approach 2nd Edition by Mohammad H. Sadraey , ISBN: 9781032245157 |All Chapters Verified| Guide A+

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Solutions Manual for Aircraft Performance: An Engineering Approach 2nd Edition by Mohammad H. Sadraey , ISBN: 9781032245157 |All Chapters Verified| Guide A+

Institution
Aircraft Performance: An Engineering Approach
Course
Aircraft Performance: An Engineering Approach











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Institution
Aircraft Performance: An Engineering Approach
Course
Aircraft Performance: An Engineering Approach

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Uploaded on
January 3, 2026
Number of pages
455
Written in
2025/2026
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Exam (elaborations)
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SOLUTION MANUAL
Aircraft Performance: An Engineering Approach

Mohammad H. Sadraey

2nd Edition
PR
O
FD
O
C



1

,Table of Contents — Aircraft Performance: An Engineering
Approach (2nd Edition)


1. Atmosphere
2. Equations of Motion
3. Drag Force and Drag Coefficient
4. Engine Performance
5. Straight-Level Flight – Jet Aircraft
6. Straight-Level Flight: Propeller-Driven Aircraft
7. Climb and Descent
8. Takeoff and Landing
PR
9. Turn Performance and Flight Maneuvers
10.Aircraft Performance Analysis Using Numerical
Methods and MATLAB
O
FD
O
C



2

, Solutions to problems for
Aircraft Performance: An Engineering Approach, Mohammad Sadraey, 2nd ed.

Ch. 1

The software package Mathcad is used to solve problems.



1.1. Determine the temperature, pressure and air density at 5,000 m and ISA condition.

There are two methods:
a. Using appendix:
From Appendix A:
PR
- Temperature: 255.69 K
- Pressure: 54,048 Pa
- Air density: 0.7364 kg/m3

b. Calculations:
O
K J
h = 5000m ISA L1 = 6.5 R1 = 287 Po = 101325Pa
1000m kgK
FD
Sea level: To = (15 + 273)K = 288 K


5000 m: T5 = To − L1h = 255.5 K (Equ 1.6)


5.256
 T5 
O
P5 = Po  = 54000.3 Pa (Equ 1.16)
 To 
C
P5 kg
5 = = 0.736 (Equ 1.23)
R1T5 3
m


Same results.




3

, 1.2. Determine the pressure at 5,000 m and ISA-10 condition.


K J
h = 5000m ISA − 10 L1 = 6.5 R1 = 287 Po = 101325Pa
1000m kgK

Sea level: To = (15 + 273 − 10)K = 278 K


5000 m: T5 = To − L1h = 245.5 K (Equ 1.6)


5.256
 T5 
P5 = Po  = 52714.2 Pa (Equ 1.16)
 To 
PR

1.3. Calculate air density at 20,000 ft altitude and ISA+15 condition.



K J
O
h = 20000ft ISA + 15 L1 = 2 R1 = 287 Po = 101325Pa
1000ft kgK

Sea level: To = [(15 + 273) + 15]K = 303 K To = 545.4R
FD

20000 ft: T20 = To − L1h = 263 K T20 = 473.4R (Equ 1.6)


5.256
 T20  lbf
O
P20 = Po  = 48143.9 Pa P20 = 1005.5 (Equ 1.16)
 To  ft
2
C
P20 kg slug
20 = = 0.638 20 = 0.001238 (Equ 1.23)
R1T20 3 3
m ft




1.4. An aircraft is flying at an altitude at which its temperature is -4.5 oC. Calculate:


4

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